Posted by Capacino on Sunday, June 21, 2009 at 11:06am.
Long but quick chemistry lab that I need confirmation on answers?
OK so I have this lab that went pretty quick and smooth and I have an online assessment that I needed to put all my figures in. I am all complete but am still having trouble with 1 problems that the assessment is saying is wrong and I am looking for someone to verify or explain what I am doing incorrect.
Here were the steps of the lab and my findings:
Determining the molecular weight of Acetone:
Step 1: Obtain a Dumas bulb from the Equipment menu.
Step 2: Record the volume of the bulb in liters (volume is marked on bulb). .056L
Step 3: Record the mass the empty bulb using an electronic balance. 50.g
Step 4: Place 20 ml of acetone into the bulb.
Step 5: Obtain a 600-ml beaker.
Step 6: Select Beaker and place about 500 ml of water into beaker or enough to cover neck of bulb.
Step 7: Combine bulb with beaker (using combine menu option).
Step 8: Obtain a hot plate from Equipment menu.
Step 9: Place combined Dumas bulb and beaker assembly onto hot plate.
Step 10: Turn on hot plate to maximum heating setting.
Step 11: Allow all acetone to boil away and wait 1-2 minutes before removing bulb from beaker.
Step 12: After removing bulb from beaker, allow Dumas bulb to cool until acetone vapors have condensed (2-3 minutes).
Step 13: Weigh bulb and condensed vapor using an electronic scale and record result. 50.1062g
Then the lab told me to use this info to answer the assessment:
(i) Volume of Dumas bulb (L): 56ml or .056L
(ii) Mass of empty Dumas bulb (g): 50g
(iii) Mass of Dumas bulb + condensed acetone (g):50.106
(iv) Mass of acetone gas required to fill bulb ((iii) - (ii)): .106
(v) Molecular weight of Acetone: Mass / n = Mass / (PV/RT) = (Mass *R *T) /(P * V) =.00183
Given:
Pressure: 1atm
Temperature: 373 Kelvin (boiling point of water)
Then Here was the question I needed to answer (which I am getting number 5 incorrect):
1)Weight of empty Dumas flask=50g
2)Weight of acetone=? 15.714
3)Weight of liquid vapor after condensing=.106
4)Moles of acetone vapor based on ideal gas law=.00183
5)Calculated molecular weight of acetone based on experiment. 58
6)%error from the theoretical molecular of acetone and the experimental value.=which to me seems like there is a 0% error
- chemistry - bobpursley, Sunday, June 21, 2009 at 11:10am
I think I would take the calculation on 5 to three significant places, since your fraction .106/.00183 indicates that.
- chemistry - Capacino, Sunday, June 21, 2009 at 12:34pm
I just cant figure out the theoretical yield???
- chemistry - DrBob222, Sunday, June 21, 2009 at 1:31pm
There is no theoretical yield which may explain why you can calculate it. There is a theoretical molar mass for acetone. That is (CH3)2CO and you can calculate the molar mass easily. Its about 58 or so but you need to do it exactly.
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